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PHY 231
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The problem:
A ball is moving at 10 cm/s when clock time is 4 seconds, and at 40 cm/s when clock time is 9 seconds.
Sketch a v vs. t graph and represent these two events by the points (4 sec, 10 cm/s) and (9 s, 40 cm/s).
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
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Sketch a straight line segment between these points.
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
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What are the rise, run and slope of this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
Rise: 30 cm/s
Run: 5 s
Slope: 6 cm/s^2
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@& Good.*@
What is the area of the graph beneath this segment?
answer/question/discussion: ->->->->->->->->->->->-> (start in the next line):
(9s+4s)/2=6.5 s
(40 cm/s+10cm/s)/2=25 cm/s
6.5 s*25 cm/s=162.5 cm
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@& If the 5-second interval ran from 20 sec to 25 sec (as would happen if the clock measuring the motion had been started 16 seconds earlier), your calculation sequence would start with (20 sec + 25 sec) / 2 = 22.5 sec, and your distance prediction would be 22.5 sec * 25 cm/s = 510 cm, approximately.
Clearly the result of moving at an average velocity of 25 cm/s for a given interval shouldn't depend on when the clock was started.
Can you see the error and correct it?*@
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